Relation between the dimensions of the fixed point and the submodule given by the augmentation ideal Let $k$ be a field with characteristic $p>0$, let $G$ be a finite $p$-group of order $n=p^m$, and let $W$ be a (left) $k[G]$-module such that the $\dim_k(W)=n$. If we denote by $I$ the augmentation ideal of $k[G]$, then we can consider $IW$ and $W^G$, two $k$-submodules of $W$.
My question is: Are $\dim_k(W/IW)$ and $\dim_k(W^G)$ equal? Or, if not equal, at least related?
My attempts:

*

*I have found some specific example that can emerge in number theory (so with Galois extension of $p$-adic fields) where this works, but the methods are not easily generalisable.

*In the general case, my first try was to look if $W^G$ and $IW$ are in direct sum, and their sum gives all $W$, but this is not true: the trace of $x\in W$ is both in $IW$ and $W^G$, and not necessarily zero.

*Maybe something cohomological? Something like $W^G=H^0(k[G],W)$ and $W/W=H_0(k[G],W)$ (we are extending the usual group cohomology in the category of (left) $k[G]$-modules, for example taking the functors Ext and Tor). But even in the classical case, where we take $Z[G]$ instead of $k[G]$, the result does not hold. (It holds, for example, if $W$ is induced). In general, I did not find useful sequences involving both homology and cohomology (except for the sequence with Tate groups).

So I am out of ideas...
Remark: if we need this, we can assume that $W=M/PM$, where $M$ is an $R$-module, for $R$ a dvr with residue field $R/P=k$, and that $\dim_k(W^G)=1$. In this case, it should be true that $\dim_k(W/IW)=1$, but I cannot see why also in this easier case.
 A: First of all, $W/IW$ is just $W_G$, the coinvariants of $W$. The modules $W^G$ and $W_G$ are not that related unless you are in a very special situation, say when $G$ is a cyclic group.
If you don't impose any restriction on $\mathrm{dim}(W)$, you can first replace $G$ by taking $G \times P$ for a random $p$-group $P$ which acts trivially on $W$ (thus making $W$ very small compared to $n$), and then increasing the size of $W$ by taking the direct product of $W$ with a trivial $G$-module. Hence putting some condition on $\mathrm{dim}(W)$ with respect to $|G|$  is not really a restriction.
(Ah, but $\mathrm{dim}(W^G) = 1$ and $\mathrm{dim}(W)=n$ are together a strong assumption, more on that later)
Here is an example to show that $\mathrm{dim}(W_G)$ and $\mathrm{dim}(W^G)$ are not the same in general.
Let $G = (\mathbf{Z}/p \mathbf{Z})^2 = \langle x,y | x^p,y^p, [x,y] \rangle$, and let $W = (\mathbf{F}_p)^3 = \{e_1,e_2,e_3\}$ be a vector space on which $G$ acts as follows:
$$x \mapsto \left( \begin{matrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 
\end{matrix} \right),$$
$$y \mapsto \left( \begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 
\end{matrix} \right).$$
Now $W^G = \{e_1\}$ has dimension $1$, whereas $W/IW=W_G = W/e_1$ has dimension $2$. This example even has the form given at the end of your post: $\mathrm{dim}(W^G) = 1$, and  $W=M/PM$ where $M$ is a $R = \mathbf{Z}_p[\zeta_p]$-module, where $P = (\zeta-1)$ the action of $G$ on $M$ is given by
$$x \mapsto \left( \begin{matrix} 1 & 1 & 0 \\ 0 & \zeta & 0 \\ 0 & 0 & 1 
\end{matrix} \right),$$
$$y \mapsto \left( \begin{matrix} \zeta & 0 & 1 \\ 0 & \zeta & 0 \\ 0 & 0 & 1 
\end{matrix} \right).$$
(Exercise: they both commute and have order $p$.)

On the other hand:

*

*Assume that $\mathrm{dim}(W) = n$

*Assume that $\mathrm{dim}(W^G)=1$.

For any
$W$, one can form $W^{\vee} = \mathrm{Hom}(W,k)$ and this is a $G$-module such that $\mathrm{dim}(W^G) = \mathrm{dim}(W^{\vee}_G)$.  The assumption that $W^G$ has dimension one implies that $W^{\vee}_G$ has dimension one. By Nakayama's lemma, it follows that $W^{\vee}_G$  is a cyclic module (if $G$ is a $p$-group then $k[G]$ is local with maximal two-sided ideal $I$). A cyclic module is a quotient of $k[G]$, so by dimension considerations $W^{\vee} = k[G]$. The the question becomes whether $k[G]^G$ is trivial for a $p$-group $G$. But it is clear that the only invariant element in $k[G]$ a multiple of $\sum_{g \in G}[g]$, so under all these assumptions it follows that $W$ and $W^{\vee}$ are both actually free of rank one over $k[G]$.
